Question

We are solutions made and the only one I did not keep was

Solution A Pure solvent A
Add 100.0 mL of distilled water to the plastic cup “A”.

Is this experiemnt instructor wanted to know the molarity in colunm 1 of solutions A=F, so me the molarity has nothing to do with the problem, it is mass, inifial mass. Can anyone elaborate.

Part E: Osmosis

This was a celery experiment, and we had solutions

7. Complete the table (24 pts: 1 pts per entry)

Copy the molality values of each solution (from Question 1 and Question 5).

Calculate the difference and % difference in the mass of the celery stalk for each solution.  

                 Mass difference = mass (after) − mass (initial)            (negative for loss; positive for gain)

                       %difference ₌^{massdifference}_×100%                                  (negativeforloss; postivie for gain)

mass(initial)

Indicate if the solution was hypertonic, hypotonic, or isotonic.

Solution	Molarity	Mass       difference	% difference	Tonicity of solution
A	2m	4.12g = .04g	Hypotonic
B	0.4m	3.81g =-0.66g	Hypertonic
C	0.2m	4.29g =-0.36g	Hypertonic
D	0.3m	4.15g =-0.4g	Hypertonic
E	1.1m	3.73g=-0.26g	Hypotpnic
F	0.	3.30g=0.10g	Hypertonic

Initial Mass

4.04g

4.47g

4.92g

4.55g

3.99g

3.40g

Show you work here for calculating mass difference and % difference.

Calculations for Solution A:

Initial mass 4.04g

Mass at 2 hours 4.12g

Mass difference = mass (after) – mass (initial) = 4.12-4.04=0.08g

% difference = mass diff/mass ini x 100% = 0.08/4.04 x 100 = 1.98%

Tonicity of solution is Hypotonic

Answer 1

Solutions	Molality	Initial mass(mi)	Final mass (mf)	Mas difference (∆m)	% difference= (∆m/mi)100	Tonicity
A	2m	4.04g	4.12g	0.08g	1.98%	Hypotonic
B	0.4m	4.47g	3.81g	- 0.66g	14.77%	Hypertonic
C	0.2m	4.92g	4.29g	- 0.63g	12.80%	Hypertonic
D	0.3m	4.55g	4.15g	- 0.4g	8.79%	Hypertonic
E	1.1m	3.99g	3.73g	- 0.26g	6.51%	Hypotonic
F	0.	3.40g	3.30g	0.10g	2.94%	Hypertonic